# How many electrons in an atom can have #n + l =6# ?

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The answer is 18. How do you get this answer?? Thanks!

The answer is 18. How do you get this answer?? Thanks!

##### 1 Answer

#### Explanation:

As you know, we use **four quantum numbers** to describe the *location* and *spin* of an electron in an atom.

In your case, you must find the number of electrons that can have

#n + l = 6" " " "color(orange)("(*)")#

The *angular momentum quantum number*, **subshell** in which the electron is located, depends on the value of the *principal quantum number*, **energy level** on which the electron resides, as given by

#l <= n-1#

Right from the start, you can tell that the first energy level that can hold electrons that satisfy condition

#n = 3 implies l = (0, 1, 2}#

you can't have a pair that matches the given condition

#3 + 0 < 6" "color(red)(xx)#

#3 + 1 < 6" "color(red)(xx)#

#3 + 2 < 6 " "color(red)(xx)#

The same is true, of course, for

Now, for

#n = 4 implies l = {0, 1, 2, 3}#

#4 + 2 = 6" "color(green)(sqrt())#

For the angular momentum quantum number, you have

#l=0 -># thes-subshell#l=1 -># thep-subshell#l=2 -># thed-subshell

The **d-subshell** holds a total of **orbitals** as given by the *magnetic quantum number*,

#m_l = {-2, -1, 0, 1, 2} -># for thed-subshell

Now, each *orbital* can hold a maximum of **two electrons**, which means that a total of

#5 color(red)(cancel(color(black)("d-orbitals"))) * "2 e"^(-1)/(1color(red)(cancel(color(black)("orbital")))) = "10 e"^(-)#

can have **and**

#n=5 implies l = {0, 1, 2, 3, 4}#

This time, you can have

#5 + 1 = 6" "color(green)(sqrt())#

The **p-subshell** holds a total of **orbitals** as given by

#m_l = {-1,0,1} -># for thep-subshell

Each of those orbitals can hold

#3 color(red)(cancel(color(black)("p-orbitals"))) * "2 e"^(-)/(1color(red)(cancel(color(black)("orbital")))) = "6 e"^(-)#

can have **and**

#n=6 implies l = {0,1,2,3,4,5}#

This time, you have

#6 + 0 = 6" "color(green)(sqrt())#

The **s-subshell** holds a *single orbital*, since

#m_l = 0 -># for thes-subshell

This means that only

#1 color(red)(cancel(color(black)("s-orbital"))) * "2 e"^(-)/(1color(red)(cancel(color(black)("orbital")))) = "2 e"^(-)#

can have **and**

Therefore, the **total number** of electrons that can have

#overbrace("10 e"^(-))^(color(blue)(4 + 2 = 6)) + overbrace("6 e"^(-))^(color(darkgreen)(5 + 1 = 6)) + overbrace("2 e"^(-))^(color(purple)(6 + 0 = 6)) = "18 e"^(-)#